Chapter 4: Differentation

### Orthogonal Coordinates

An orthogonal coordinate system is one in which the coordinate directions are mutually perpendicular. The standard examples are rectangular, polar, cylindrical and spherical coordinates.

Working in Euclidean $\RR^3$, suppose that $(u,v,w)$ are orthogonal coordinates. Then an infinitesimal displacement $d\rr$ between nearby points can be expressed as $$d\rr = h_u\,du \,\Hat u + h_v\,dv \,\Hat v + h_w\,dw \,\Hat w \label{ortho}$$ for some functions $h_u$, $h_v$, $h_w$, where $\{\Hat u,\Hat v,\Hat w\}$ is an orthonormal basis. Equivalently, the line element takes the form $$ds^2 = d\rr \cdot d\rr = h_u^2 \,du^2 + h_v^2 \,dv^2 + h_w^2 \,dw^2$$ Since $d\rr$ expresses the infinitesimal displacement in mutually orthogonal directions, its components, namely $\{h_u\,du,h_v\,dv,h_w\,dw\}$, should be an orthonormal basis of 1-forms. 1)

We also have $$d\rr = \Partial{\rr}{u}\,du + \Partial{\rr}{v}\,dv + \Partial{\rr}{w}\,dw$$ and comparison with ($\ref{ortho}$) shows that $$h_u\,\Hat u = \Partial{\rr}{u}$$ which in turn implies that \begin{align} h_u &= \left|\Partial{\rr}{u}\right| \\ \Hat u &= \frac{1}{h_u}\Partial{\rr}{u} \end{align} and similarly for $v$ and $w$. Thus, if $\rr$ can be expressed in terms of $u,v,w$ — the traditional parameterization — then the orthonormal bases (of both vectors and 1-forms) can be easily computed.

1) This could be verified by expressing $u,v,w$ in terms of $x,y,z$, substituting the result into $ds^2$, and comparing with the known result $ds^2 = dx^2 + dy^2 + dz^2$. The resulting constraints can be solved and inserted into $du$, $dv$, $dw$, expressed in terms of $dx$, $dy$, $dz$, and $g$ can now be evaluated in this basis. This method is however equivalent to inverting a $3\times3$ matrix; we will learn an easier method in §Vectors and Differential Forms.